\(\int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} (-1-x^4+x^8)} \, dx\) [2564]

Optimal result

Integrand size = 34, antiderivative size = 217 \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right )-\frac {1}{2} \sqrt {\frac {1}{10} \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-1+x^4}}\right ) \]

[Out]

arctan(x/(x^4-1)^(1/4))-1/20*(-10+10*5^(1/2))^(1/2)*arctan(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4-1)^(1/4))-1/20*(10+
10*5^(1/2))^(1/2)*arctan(1/2*(2+2*5^(1/2))^(1/2)*x/(x^4-1)^(1/4))+arctanh(x/(x^4-1)^(1/4))-1/20*(-10+10*5^(1/2
))^(1/2)*arctanh(1/2*(-2+2*5^(1/2))^(1/2)*x/(x^4-1)^(1/4))-1/20*(10+10*5^(1/2))^(1/2)*arctanh(1/2*(2+2*5^(1/2)
)^(1/2)*x/(x^4-1)^(1/4))

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {6860, 246, 218, 212, 209, 1442, 385} \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {5}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {5}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {5}} \]

[In]

Int[(-1 - 2*x^4 + 2*x^8)/((-1 + x^4)^(1/4)*(-1 - x^4 + x^8)),x]

[Out]

ArcTan[x/(-1 + x^4)^(1/4)] - (((3 - Sqrt[5])/2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*x)/(-1 + x^4)^(1/4)])/(2
*Sqrt[5]) - (((3 + Sqrt[5])/2)^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1/4)*x)/(-1 + x^4)^(1/4)])/(2*Sqrt[5]) + ArcTa
nh[x/(-1 + x^4)^(1/4)] - (((3 - Sqrt[5])/2)^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*x)/(-1 + x^4)^(1/4)])/(2*Sq
rt[5]) - (((3 + Sqrt[5])/2)^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*x)/(-1 + x^4)^(1/4)])/(2*Sqrt[5])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{\sqrt [4]{-1+x^4}}+\frac {1}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt [4]{-1+x^4}} \, dx+\int \frac {1}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\frac {2 \int \frac {1}{\sqrt [4]{-1+x^4} \left (-1-\sqrt {5}+2 x^4\right )} \, dx}{\sqrt {5}}-\frac {2 \int \frac {1}{\sqrt [4]{-1+x^4} \left (-1+\sqrt {5}+2 x^4\right )} \, dx}{\sqrt {5}} \\ & = \frac {2 \text {Subst}\left (\int \frac {1}{-1-\sqrt {5}-\left (1-\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {5}}-\frac {2 \text {Subst}\left (\int \frac {1}{-1+\sqrt {5}-\left (1+\sqrt {5}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {5}}+\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right ) \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {10}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {10}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {10}}-\frac {\text {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt {10}} \\ & = \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{-1+x^4}}\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {5}}+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} x}{\sqrt [4]{-1+x^4}}\right )}{2^{3/4} \sqrt {5} \sqrt [4]{3+\sqrt {5}}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {5}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.93 \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\frac {1}{20} \left (20 \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {10 \left (-1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {10 \left (1+\sqrt {5}\right )} \arctan \left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )+20 \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {10 \left (-1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )-\sqrt {10 \left (1+\sqrt {5}\right )} \text {arctanh}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt [4]{-1+x^4}}\right )\right ) \]

[In]

Integrate[(-1 - 2*x^4 + 2*x^8)/((-1 + x^4)^(1/4)*(-1 - x^4 + x^8)),x]

[Out]

(20*ArcTan[x/(-1 + x^4)^(1/4)] - Sqrt[10*(-1 + Sqrt[5])]*ArcTan[(Sqrt[(-1 + Sqrt[5])/2]*x)/(-1 + x^4)^(1/4)] -
Sqrt[10*(1 + Sqrt[5])]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/(-1 + x^4)^(1/4)] + 20*ArcTanh[x/(-1 + x^4)^(1/4)] -
Sqrt[10*(-1 + Sqrt[5])]*ArcTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/(-1 + x^4)^(1/4)] - Sqrt[10*(1 + Sqrt[5])]*ArcTanh[
(Sqrt[(1 + Sqrt[5])/2]*x)/(-1 + x^4)^(1/4)])/20

Maple [A] (verified)

Time = 3.01 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {\sqrt {2+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )}{40}-\frac {\sqrt {-2+2 \sqrt {5}}\, \left (5+\sqrt {5}\right ) \operatorname {arctanh}\left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )}{40}-\frac {\sqrt {2+2 \sqrt {5}}\, \left (-5+\sqrt {5}\right ) \arctan \left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {-2+2 \sqrt {5}}\, x}\right )}{40}+\frac {\sqrt {-2+2 \sqrt {5}}\, \left (5+\sqrt {5}\right ) \arctan \left (\frac {2 \left (x^{4}-1\right )^{\frac {1}{4}}}{\sqrt {2+2 \sqrt {5}}\, x}\right )}{40}+\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}+x}{x}\right )}{2}-\arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )-\frac {\ln \left (\frac {-x +\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{2}\)	\(202\)

[In]

int((2*x^8-2*x^4-1)/(x^4-1)^(1/4)/(x^8-x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/40*(2+2*5^(1/2))^(1/2)*(-5+5^(1/2))*arctanh(2/(-2+2*5^(1/2))^(1/2)/x*(x^4-1)^(1/4))-1/40*(-2+2*5^(1/2))^(1/2
)*(5+5^(1/2))*arctanh(2/(2+2*5^(1/2))^(1/2)/x*(x^4-1)^(1/4))-1/40*(2+2*5^(1/2))^(1/2)*(-5+5^(1/2))*arctan(2/(-
2+2*5^(1/2))^(1/2)/x*(x^4-1)^(1/4))+1/40*(-2+2*5^(1/2))^(1/2)*(5+5^(1/2))*arctan(2/(2+2*5^(1/2))^(1/2)/x*(x^4-
1)^(1/4))+1/2*ln(((x^4-1)^(1/4)+x)/x)-arctan((x^4-1)^(1/4)/x)-1/2*ln((-x+(x^4-1)^(1/4))/x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (145) = 290\).

Time = 0.26 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.87 \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=-\frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} + 1} + 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} + 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} + 1} - 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} - 1} + 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {\sqrt {5} - 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {\sqrt {5} - 1} - 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} + 1} + 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} + 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} + 1} - 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} - 1} + 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{40} \, \sqrt {10} \sqrt {-\sqrt {5} - 1} \log \left (-\frac {\sqrt {10} \sqrt {5} x \sqrt {-\sqrt {5} - 1} - 10 \, {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate((2*x^8-2*x^4-1)/(x^4-1)^(1/4)/(x^8-x^4-1),x, algorithm="fricas")

[Out]

-1/40*sqrt(10)*sqrt(sqrt(5) + 1)*log((sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) + 1) + 10*(x^4 - 1)^(1/4))/x) + 1/40*sqr
t(10)*sqrt(sqrt(5) + 1)*log(-(sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) + 1) - 10*(x^4 - 1)^(1/4))/x) - 1/40*sqrt(10)*sq
rt(sqrt(5) - 1)*log((sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) - 1) + 10*(x^4 - 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(sqrt(5
) - 1)*log(-(sqrt(10)*sqrt(5)*x*sqrt(sqrt(5) - 1) - 10*(x^4 - 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(-sqrt(5) + 1)*
log((sqrt(10)*sqrt(5)*x*sqrt(-sqrt(5) + 1) + 10*(x^4 - 1)^(1/4))/x) - 1/40*sqrt(10)*sqrt(-sqrt(5) + 1)*log(-(s
qrt(10)*sqrt(5)*x*sqrt(-sqrt(5) + 1) - 10*(x^4 - 1)^(1/4))/x) + 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log((sqrt(10)
*sqrt(5)*x*sqrt(-sqrt(5) - 1) + 10*(x^4 - 1)^(1/4))/x) - 1/40*sqrt(10)*sqrt(-sqrt(5) - 1)*log(-(sqrt(10)*sqrt(
5)*x*sqrt(-sqrt(5) - 1) - 10*(x^4 - 1)^(1/4))/x) - arctan((x^4 - 1)^(1/4)/x) + 1/2*log((x + (x^4 - 1)^(1/4))/x
) - 1/2*log(-(x - (x^4 - 1)^(1/4))/x)

Sympy [F(-1)]

Timed out. \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\text {Timed out} \]

[In]

integrate((2*x**8-2*x**4-1)/(x**4-1)**(1/4)/(x**8-x**4-1),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - 2 \, x^{4} - 1}{{\left (x^{8} - x^{4} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((2*x^8-2*x^4-1)/(x^4-1)^(1/4)/(x^8-x^4-1),x, algorithm="maxima")

[Out]

integrate((2*x^8 - 2*x^4 - 1)/((x^8 - x^4 - 1)*(x^4 - 1)^(1/4)), x)

Giac [F]

\[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - 2 \, x^{4} - 1}{{\left (x^{8} - x^{4} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

integrate((2*x^8-2*x^4-1)/(x^4-1)^(1/4)/(x^8-x^4-1),x, algorithm="giac")

[Out]

integrate((2*x^8 - 2*x^4 - 1)/((x^8 - x^4 - 1)*(x^4 - 1)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1-2 x^4+2 x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx=\int \frac {-2\,x^8+2\,x^4+1}{{\left (x^4-1\right )}^{1/4}\,\left (-x^8+x^4+1\right )} \,d x \]

[In]

int((2*x^4 - 2*x^8 + 1)/((x^4 - 1)^(1/4)*(x^4 - x^8 + 1)),x)

[Out]

int((2*x^4 - 2*x^8 + 1)/((x^4 - 1)^(1/4)*(x^4 - x^8 + 1)), x)